For which real $a$ is the integral $\int_1^\infty x^ae^{-x^3\sin^2x}dx$ finite? For which $a \in \mathbb{R}$ is the integral $\int_1^\infty x^ae^{-x^3\sin^2x}dx$ finite?
I've been struggling with this question. Obviously when $a<-1$ the integral converges, but I have no idea what happens when $a\ge -1 $.
Any help would be appreciated
 A: If $a < +0.5$, the integral converges. This is because we can split the integral over $[1,\infty)$ into an integral over $\bigcup_k (-\epsilon_k+k\pi,+\epsilon_k+k\pi)$ and an integral over all the rest.
The integral over $(-\epsilon_k + k\pi, +\epsilon_k+k\pi)$ can be bounded by $(k\pi)^a \cdot 2\epsilon_k$. Therefore: If we choose $\epsilon_k:=k^{-1.5+\delta}$ then $\epsilon_k \to 0$ fast enough such that the sum of the integrals over all the little bits will converge.
Now for the complement of the little bits. Outside of $(-\epsilon_k,+\epsilon_k)+k\pi$ we have $\sin(x)^2\geq \sin(k^{-1.5+\delta})^2$ and thus $-x^3 \sin(x)^2 \leq -k^{2\delta} \pi^3 (k^{1.5-\delta}\sin(k^{-1.5+\delta}))^2$ and $k^{1.5-\delta}\sin(k^{-1.5+\delta}) \xrightarrow{k\to\infty} 1$ because $\lim_{h\to 0} \frac{\sin(h)}{h}=1$. Therefore the integral over the complement of the little bits can be bounded from above by $\int_1^\infty x^a e^{-k^{2\delta}C}$ for some constant $C>0$. And this integral converges for all $\delta>0$.

Conversely: If $a\geq 0.5$, the integral diverges. Again we look at the integrals over the little bits. This time we choose $\epsilon_k := k^{-1.5}$. Then we use $|\sin(x)|\leq |x|$ for $x$ near zero and get
$\int_{-\epsilon_k+k\pi}^{+\epsilon_k+k\pi} x^a e^{-x^3 \sin(x)^2} \geq \int x^a e^{-x^3 \epsilon_k^2} \geq 2\epsilon_k \cdot (-\epsilon_k+k\pi)^a e^{-(\epsilon_k+k\pi)^3 k^{-3}}$
This behaves asymptotically like $k^a$ because the term in the exponent goes to $1$. Therefore the sum over all these integrals goes to infinity.
A: Fix $0<\delta <\pi/2$.  We can write the integral of interest $I(a)$ as
$$\begin{align}
I(a)&=\int_1^\infty x^ae^{-x^3\sin^2(x)}\\\\
&=\int_1^{\pi-\delta} x^a e^{-x^3\sin^2(x)}\,dx+\sum_{n=1}^\infty \int_{n\pi-\delta}^{n\pi+\delta}x^a e^{-x^3\sin^2(x)}\,dx+\sum_{n=1}^\infty \int_{n\pi+\delta}^{(n+1)\pi-\delta}x^a e^{-x^3\sin^2(x)}\,dx \tag 1
\end{align}$$
The term in $(1)$ that requires consideration is the series $\sum_{n=1}^\infty \int_{n\pi-\delta}^{n\pi+\delta}x^a e^{-x^3\sin^2(x)}\,dx$ since $\sin(x)=0$ for $x=n\pi$ and the exponential becomes $1$ there.  
Note that we can write
$$\int_{n\pi-\delta}^{n\pi+\delta}x^a e^{-x^3\sin^2(x)}\,dx=\int_{-\delta}^\delta (x+n\pi)^a e^{-(x+n\pi)^3\sin^2(x)} \tag 2$$
Heuristically, for "small" $\delta$, we can approximate the integral in $(2)$ as 
$$\begin{align}
\int_{-\delta}^\delta (x+\pi)^a e^{-(x+n\pi)^3\sin^2(x)}\,dx&\approx (n\pi)^a \int_{-\delta}^\delta  e^{-(n\pi)^3x^2}\,dx\\\\
&=(n\pi)^{a-3/2} \int_{-\delta(n\pi)^{3/2}}^{\delta (n\pi)^{3/2}}  e^{-x^2}\,dx\\\\
&\le (n\pi)^{a-3/2}\sqrt \pi
\end{align}$$
Inasmuch as the series $\sum_{n=1}^{\infty}\frac{1}{n^{3/2-a}}$ converges for values of $a<1/2$, we conclude that the integral of interest $I(a)$ converges for values of $a<1/2$.
